Question

How is momentum conserved in this electromagnetic scattering?

Answer 1

I have been pondering over an energy conservation issue for a few weeks and would like to share it with the hope that its interesting and someone will be able to answer.

Suppose I have two positive charges at (−a,0,0)and (a,0,0) connected by a spring in such a way the the above positions are equilibrium positions. Now suppose I shoot a massless positively charged particle moving along the y axis in such a way that it passes the x-axis at time t=0.

According to Jackson problem 11.18 the electric and magnetic fields of the massless particle exist only in the x-z plane passing through the instantaneous location of the particle and are given by

→E=2q→r⊥|r⊥|2δ(ct−z)

and

→B=2q→v×→r⊥|r⊥|2δ(ct−z)where v is the 3-velocity of the particle so in our case (0,0,1)

Thus, while there is a force on the massless particle when approaching the x-z plane, there is no force by the massless charge on the massive charges connected by springs except for an impulse at t=0. This impulse gives an equal and opposite momentum to both the charged particle.

|p|=2qa

When the massless charge moves away from the x-z plane the electric and magnetic fields it "feels" is equal and opposite to what it felt when approaching because any change produced by the massive charges moving will not reach the massless particle. Thus the energy in the massless particle is the same at very early and very late times.

We are left with the two massive charges oscillating. Where does the energy come from? It is tempting to say the energy comes from the field but how? At the end of the day, due to radiation losses the charged particles will go back to being at rest at (−a,0,0) and (a,0,0) and the only difference would be the field configuration produced by them. The massless particle seems to have just come along for a ride.

Is there a true degeneracy in that there are two field configurations with charges at (−a,0,0) and (a,0,0) with the exact same energy

Answer 2

It is well known that verifying momentum and energy conservation in the presence of electromagnetic fields requires care as the fields themselves carry energy and momentum (see Griffiths chapter 8 for instance). However, I have a thought experiment where I cannot seem to 'balance out' Newton's third law as I don't see how to give the electromagnetic field any compensating momentum. Thoughts and answers welcome.

Suppose I have a uniformly charged infinite plane placed on the plane z=0z=0. We can take the charge per unit area to be σσ. This gives a uniform electric field

|Ez|=σ2 ,|Ez|=σ2 ,

pointing away from the plate. Now lets consider a massless positively charged particle approaching the plate from the negative zz direction with instantaneous position

r⃗ p=(0,0,ct) ,r→p=(0,0,ct) ,

such that it will cross the plate at time t=0t=0.

According to Jackson 3rd edition problem 11.18 the electric and magnetic fields of the massless particle exist only in the x-y plane passing through the instantaneous location of the particle and are given by

E⃗ =2qr⃗ ⊥|r⊥|2δ(ct−z) ,E→=2qr→⊥|r⊥|2δ(ct−z) ,

and

B⃗ =2qv⃗ ×r⃗ ⊥|r⊥|2δ(ct−z) ,B→=2qv→×r→⊥|r⊥|2δ(ct−z) ,

where v is the 3-velocity of the particle so in our case (0,0,1)(0,0,1) and r⃗ ⊥r→⊥ is the location of the observation point in the z=ctz=ct plane.

Thus, while there is a force on the massless particle when approaching the x-y plane, there is noforce on the charged plate. Like I said before, in electromagnetism forces between charges do not have to cancel as the field can carry momentum but in this case the Poynting vector is independent of time so its not clear how to restore momentum conservation.

For completeness let me write down the Lorentz force for massless particles. It is simply

dpαdt=qFαβdxβdt ,dpαdt=qFαβdxβdt ,

where the derivative can be taken with respect to any affine parameter so I take it to be with respect to the lab frame. In our case this simply reduces to

dEdt=−qE ,dEdt=−qE ,

where EE is the lab frame energy of the massless particle.